Problem: Let $h(x)=\begin{cases} e^{x+2}&\text{for }x \leq -2 \\\\ (x+2)^e&\text{for }x> -2 \end{cases}$ Is $h$ continuous at $x=-2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Answer: For $h$ to be continuous at $x=-2$, we need $\lim_{x\to -2}h(x)$ and $h(-2)$ to exist and be equal. Since $-2\leq -2$, the rule that applies to $x=-2$ is $e^{x+2}$. So $h(-2)=e^{(-2)+2}=1$. Now let's analyze $\lim_{x\to -2}h(x)$. Finding $\lim_{x\to -2^{ +}}h(x)$ For $x$ -values larger than $-2$, the appropriate rule for $h(x)$ is $(x+2)^e$. Since $(x+2)^e$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -2^{ +}}h(x) \\\\ &=\lim_{x\to -2^{ +}}[(x+2)^e] \gray{(x+2)^e\text{ is the rule for }x>-2} \\\\ &=(-2+2)^e \gray{(x+2)^e\text{ is continuous at }x=-2} \\\\ &=0 \end{aligned}$ Finding $\lim_{x\to -2^{ -}}h(x)$ For $x$ -values smaller than $-2$, the appropriate rule for $h(x)$ is $e^{x+2}$. Since $e^{x+2}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -2^{ -}}h(x) \\\\ &=\lim_{x\to -2^{ -}}[e^{x+2}] \gray{e^{x+2}\text{ is the rule for }x<-2} \\\\ &=e^{-2+2} \gray{e^{x+2}\text{ is continuous at }x=-2} \\\\ &=1 \end{aligned}$ Conclusion We found that $\lim_{x\to -2^{ +}}h(x)=0$ and $\lim_{x\to -2^{ -}}h(x)=1$. Since the one-sided limits aren't equal, $\lim_{x\to -2}h(x)$ doesn't exist and $h$ isn't continuous at $x=-2$.